If {a^{x - 1}} = bc,{b^{y - 1}} = ca,and {c^{ | vedclass

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(A) Given equations are: $a^{x-1} = bc \implies a^x = abc$ $b^{y-1} = ca \implies b^y = abc$ $c^{z-1} = ab \implies c^z = abc$ Taking log on both side

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(A) Given equations are: $a^{x-1} = bc \implies a^x = abc$ $b^{y-1} = ca \implies b^y = abc$ $c^{z-1} = ab \implies c^z = abc$ Taking log on both sides with base $abc$: $\log_{abc}(a^x) = \log_{abc}(abc) \implies x \log_{abc} a = 1 \implies \frac{1}{x} = \log_{abc} a$ Similarly,$\frac{1}{y} = \log_{abc} b$ and $\frac{1}{z} = \log_{abc} c$ Now,$\sum \frac{1}{x} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \log_{abc} a + \log_{abc} b + \log_{abc} c$ Using the property $\log m + \log n = \log(mn)$: $\sum \frac{1}{x} = \log_{abc} (abc) = 1$

If ${a^{x - 1}} = bc$,${b^{y - 1}} = ca$,and ${c^{z - 1}} = ab$,then find the value of $\sum \frac{1}{x}$.

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